Page 1
Forms of Energy Generation:
1. Degradation of electrical energy to heat
2. Heat from nuclear source (by fission)
3. Heat from viscous dissipation
Overall Shell Energy Balance
Energy Generation
Let S = rate of heat production per unit volume (W/m3)
(Se)
(Sn)
(Sv)
Page 2
Electrical Heat Source
Consider an electrical wire (solid cylinder):
Shell Heat Balance:
(2πππΏππ ) |π β (2πππΏππ ) |π+βπ+ (2π π βππΏ )ππ=0
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2πππΏππ ) |π(2πππΏππ ) |π+βπ
(2ππ βππΏ )ππ
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Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2πππΏππ ) |π(2πππΏππ ) |π+βπ
(2ππ βππΏ )ππ
The Shell:
(2πππΏππ ) |π=(2πππΏ) β (ππ |π )
Rate of Heat IN Area perpendicular to qr at r = r
Page 4
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2πππΏππ ) |π(2πππΏππ ) |π+βπ
(2ππ βππΏ )ππ
The Shell:
(2πππΏππ ) |π+βπ=(2π(π +βπ )πΏ) β (ππ |π +βπ )
Rate of Heat OUT Area perpendicular to qr at r = r + dr
Page 5
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(2πππΏππ ) |π(2πππΏππ ) |π+βπ
(2ππ βππΏ )ππ
The Shell:
Generation = Volume X Se
π=π [ (π+βπ )2βπ2 ] πΏToo small
β΄π=π [2π βπ ]πΏ β΄πΊππ=2π π βππΏ βππ
ΒΏπ [π2+2π βπ+ (βπ )2βπ2 ] πΏ
Page 6
Electrical Heat Source
Consider an electrical wire (solid cylinder):
Shell Heat Balance:
(2πππΏππ ) |π β (2πππΏππ ) |π+βπ+ (2π π βππΏ )ππ=0
Dividing by
(π ππ )|πβ (π ππ )|π+βπ
βπ =βππ π
Q: Why did we divide by and not by ?
(2πππΏππ ) |π β (2πππΏππ ) |π+βπ=β (2ππ βππΏ)ππ
Page 7
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:(π ππ )|πβ (π ππ )|π+βπ
β π =βππ π
Taking the limit as :
Q: Is this correct?
πππ (π ππ )=βππ π
NO!
Page 8
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:(π ππ )|πβ (π ππ )|π+βπ
β π =βππ π
We must adhere to the definition of the derivative:
(π ππ )|π+β πβ (π ππ ) |πβπ =+ππ π
limβπβ0
(π ππ )|π+β πβ (π ππ ) |πβ π = π
ππ (π ππ )=πππ
Page 9
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:πππ (π ππ )=πππ
Boundary conditions:
ππ‘ π=0 ,ππ= πππππ‘π
ππ‘ π=π
,π=π 0
Note: The problem statement will tell you hints about what boundary conditions to use.
Integrating: ππ=πππ2
+πΆ1
π
Page 10
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have:
ππ‘ π=0 ,ππ= πππππ‘πApplying B.C. 1:
ππ=πππ2
+πΆ1
π
Because q has to be finite at r = 0, all the terms with radius, r, below the denominator must vanish. Therefore:
πΆ1=0
ππ=πππ2
Page 11
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have: ππ=πππ2
Substituting Fourierβs Law:
βπ ππππ =
ππ π2
ππππ =
βπππ2π
π=βπππ 2
4π +πΆ2
Page 12
Electrical Heat Source
Consider an electrical wire (solid cylinder):
We now have: π=βπππ 2
4π +πΆ2
Applying B.C. 2: ππ‘ π=π
,π=π 0
π 0=βπππ
2
4 π +πΆ2
πΆ2=π0+πππ
2
4 π
π=βπππ 2
4π +πππ
2
4π +π 0
This is it! But, we rewrite it into a nicer formβ¦
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Electrical Heat Source
Consider an electrical wire (solid cylinder):
π βπ0=πππ
2
4π [1β( ππ
)2]
Temperature Profile:
Important assumptions:
1. Temperature rise is not large so that
k and Se are constant & uniform.
2. The surface of the wire is
maintained at T0.
3. Heat flux is finite at the center.
Page 14
Electrical Heat Source
Other important notesβ¦Let: electrical conductivity
current density
voltage drop over a length
[ 1Ξ©βππ ]
[ πππππ2 ]
[π ]
ππ=πΌ 2ππ
πΌ=πππΈπΏ
πππ
2
4π =πΈ2π
2
4πΏ2 (ππ
π )These imply the following : π βπ0=
πππ
2
4π [1β( ππ
)2]
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Electrical Heat Source
π βπ0=πππ
2
4π [1β( ππ
)2]
Temperature Profile:
The stress profile versus the temperature profile:
ππ=πππ2
Heat flux profile:
Page 16
Electrical Heat Source
Quantities that might be asked for:
1. Maximum Temperature
2. Average Temperature Rise
3. Heat Outflow Rate at the Surface
Substituting r = 0 to the profile T(r): ππππ₯=π 0+
πππ
2
4 π
β¨π β©βπ 0=β«0
2π
β«0
π
(π βπ0 )π ππ π π
β«0
2 π
β«0
π
π ππ π πβ¨π β©βπ 0=
πππ
2
8π =12ππππ₯
ππ=ππ΄= π
2π π
πΏ=πππ2
|π=π
π=π π
2πΏ βππ
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Electrical Heat Source
Examples for Review:
Example 10.2-1 and Example 10.2-2Bird, Stewart, and Lightfoot, Transport Phenomena, 2nd Ed., p. 295
Page 18
Nuclear Heat Source
Consider a spherical nuclear fuel assembly (solid sphere):
Before doing a balance, let: volumetric heat rate of production within the fissionable material only
volumetric heat rate of production at r = 0
Sn depends on radius parabolically:
a dimensionless positive constant
Page 19
Nuclear Heat Source
Consider a spherical nuclear fuel assembly (solid sphere):
Before doing a balance, let:
temperature profile in the fissionable sphere
temperature profile in the Alcladding
heat flux in the fissionable sphere
heat flux in the Al cladding
Page 20
Nuclear Heat Source
Consider a spherical nuclear fuel assembly (solid sphere):
For the fissionable material:
(4π π2ππ(πΉ ))|πβ (4ππ 2ππ
(πΉ) )|π+β π+(4 ππ2βπ )ππ=0
Rate of Heat IN:
Rate of Heat OUT:
Generation:
(4π π2ππ(πΉ ))|π
(4π π2ππ(πΉ ))|π+βπ
(4ππ 2β π )ππ
Page 21
Electrical Heat Source
Rate of Heat IN:
Rate of Heat OUT:
Generation:
Generation = Volume X Sn
π=43 π [ (π+βπ )3βπ3 ]
Too small
β΄π=4π [π2βπ ] β΄πΊππ=(4ππ 2βπ )ππ
ΒΏ43 π [π3+3π 2βπ +3π (β π )2+ (βπ )3βπ 3 ]
(4π π2ππ(πΉ ))|π
(4π π2ππ(πΉ ))|π+βπ
(4ππ 2β π )ππ
Page 22
Nuclear Heat Source
For the fissionable material:
(4π π2ππ(πΉ ))|πβ (4ππ 2ππ
(πΉ) )|π+β π+(4 ππ2βπ )ππ=0
For the Al cladding:
(4π π2ππ(πΆ ))|πβ ( 4ππ 2ππ
(πΆ))|π +βπ=0
No generation
here!
Dividing by :
(π 2ππ(πΉ ))|π+βπ β (π 2ππ
(πΉ ) )|πβπ =πππ
2
Dividing by :
(π 2ππ(πΆ))|π+βπβ (π 2ππ
(πΆ ))|πβπ =0
Page 23
Nuclear Heat Source
For the fissionable material:
(4π π2ππ(πΉ ))|πβ (4ππ 2ππ
(πΉ) )|π+β π+(4 ππ2βπ )ππ=0
For the Al cladding:
(4π π2ππ(πΆ ))|πβ ( 4ππ 2ππ
(πΆ))|π +βπ=0
No generation
here!
Taking :
πππ (π 2ππ
(πΉ ))=πππ 2
Taking :
πππ (π 2ππ
(πΆ))=0
Page 24
Nuclear Heat Source
For the fissionable material:
(4π π2ππ(πΉ ))|πβ (4ππ 2ππ
(πΉ) )|π+β π+(4 ππ2βπ )ππ=0
For the Al cladding:
(4π π2ππ(πΆ ))|πβ ( 4ππ 2ππ
(πΆ))|π +βπ=0
No generation
here!
Taking :
πππ (π 2ππ
(πΉ ))=ππ0[1+π ( ππ
(πΉ ) )
2]π 2Taking :
πππ (π 2ππ
(πΆ))=0
Page 25
Nuclear Heat Source
For the fissionable material:
(4π π2ππ(πΉ ))|πβ (4ππ 2ππ
(πΉ) )|π+β π+(4 ππ2βπ )ππ=0
For the Al cladding:
(4π π2ππ(πΆ ))|πβ ( 4ππ 2ππ
(πΆ))|π +βπ=0
No generation
here!
Integrating: Integrating:
Page 26
Nuclear Heat Source
Integrating: Integrating:
Boundary Conditions: Boundary Conditions:
ππ‘ π=0 ,ππ ππ πππππ‘π ππ‘ π=π
(πΉ ) ,ππ(πΉ )=ππ
(πΆ )
πΆ1(πΉ )=0 πΆ1
(πΆ)=ππ0 ( 13+π5 )π
(πΉ )3
For the fissionable material For the Al cladding
Page 27
Nuclear Heat Source
Inserting Fourierβs Law: Inserting Fourierβs Law:
For the fissionable material For the Al cladding
Page 28
Nuclear Heat Source
For the fissionable material For the Al cladding
Boundary Conditions: Boundary Conditions:
At r = R(F),
T(F) = T(C) R(F)
R(C)
At r = R(C),
T(C) = T0
Page 29
Nuclear Heat Source
For the fissionable material
For the Al cladding
Page 30
Recall the Overall Shell Energy Balance:
Overall Shell Energy Balance
Q by Convective Transport Q by Molecular Transport
W by Molecular Transport W by External Forces
Energy Generation
Steady-State!
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Overall Shell Energy Balance
Q by Convective Transport Q by Molecular Transport
W by Molecular Transport
How can we account for all these terms at once?
We need all these terms for viscous dissipation:
Page 32
Combined Energy Flux Vector
Convective Energy FluxHeat Rate from Molecular Motion
Work Rate from Molecular Motion
Combined Energy Flux Vector:
π=( 12 ππ£2+π οΏ½ΜοΏ½)π+ [π
βπ ]+π
We introduce something new to replace q:
Page 33
Combined Energy Flux Vector
Combined Energy Flux Vector:
We introduce something new to replace q:
π
=ππΉ+πRecall the molecular stress tensor:When dotted with v: [π
βπ ]=π π+[π βπ ]
Substituting into e:
π=( 12 ππ£2+π οΏ½ΜοΏ½)π+ππ+[π βπ ]+π
Page 34
Combined Energy Flux Vector
Combined Energy Flux Vector:
We introduce something new to replace q:
π=( 12 ππ£2+π οΏ½ΜοΏ½)π+ππ+[π βπ ]+π
Simplifying the boxed expression:
π οΏ½ΜοΏ½ π+π π=π (οΏ½ΜοΏ½+ ππ )π=π (π+π οΏ½ΜοΏ½ ) π=π οΏ½ΜοΏ½ π
Finally: π=( 12 ππ£2+π οΏ½ΜοΏ½ )π+[π βπ ]+π
Page 35
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
π£ π§ (π₯ )=π£π( π₯π )
Page 36
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
We now make a shell balance shown in red on the left.
Rate of Energy IN:
Rate of Energy OUT:
ππΏππ |π₯
ππΏππ |π₯+β π₯
When the combined energy flux vector is used, the generation term will automatically appear from e.
Page 37
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
We now make a shell balance shown in red on the left.
Rate of Energy IN:
Rate of Energy OUT:
ππΏππ |π₯
ππΏππ |π₯+β π₯
When the combined energy flux vector is used, the generation term will automatically appear from e.
ππΏππ |π₯+β π₯βππΏππ |π₯=0πππ
ππ₯ =0
ππ=πΆ1
Page 38
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
ππ=πΆ1 π=( 12 ππ£2+π οΏ½ΜοΏ½ )π+[π βπ ]+π
Fourierβs Law:
Newtonβs Law:
ππ₯=βπππππ₯
ππ₯π§=βπππ£ π§
ππ₯
Page 39
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
ππ=πΆ1 βπ ππππ₯ βππ£π§
ππ£π§
ππ₯ =πΆ1
Substituting the velocity profile:
Integrating:
βπ ππππ₯ βππ₯ (π£π
π )2
=πΆ1
π=β ππ ( π£π
π )2 π₯22βπΆ1
π π₯+πΆ2
Page 40
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
Boundary Conditions:
After applying the B.C.:
π βπ 0
ππβπ0=12 [ ππ£π
2
π (π πβπ 0 ) ]( π₯π )(1β π₯π )+π₯π
π=β ππ ( π£π
π )2 π₯22βπΆ1
π π₯+πΆ2
Page 41
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
When the combined energy flux vector is used, the generation term will automatically appear from e.
π βπ 0
ππβπ0=12 [ ππ£π
2
π (π πβπ 0 ) ]( π₯π )(1β π₯π )+π₯π
Q: So where is Sv?
ππ£=π(π£π
π )2
π=β ππ ( π£π
π )2 π₯22βπΆ1
π π₯+πΆ2
After applying the B.C.:
Page 42
Viscous Dissipation Source
Consider the flow of an incompressible Newtonian fluid between 2 coaxial cylinders:
π βπ 0
ππβπ0=12 [ ππ£π
2
π (π πβπ 0 ) ]( π₯π )(1β π₯π )+π₯πTemperature
Profile:
New Dimensionless Number:Dim. Group Ratio Equation
Brinkman, Br viscous heat dissipation/ molecular heat transport
Page 43
Viscous Dissipation Source
Scenarios when viscous heating is significant:
1. Flow of lubricant between rapidly moving parts.2. Flow of molten polymers through dies in high-
speed extrusion.3. Flow of highly viscous fluids in high-speed
viscometers.4. Flow of air in the boundary layer near an earth
satellite or rocket during reentry into the earthβs atmosphere.